Problem: What is the average value of $\sqrt{x}$ on the interval $[5,12]$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{2}{21}\cdot (\sqrt{12^3}-\sqrt{5^3})$ (Choice B) B $\dfrac{2}{21}\cdot (\sqrt[3]{12^2}-\sqrt[3]{5^2})$ (Choice C) C $\dfrac{1}{2}\cdot (\sqrt{12}-\sqrt{5})$ (Choice D) D $\dfrac{1}{2}\cdot (\sqrt{12}+\sqrt{5})$
Explanation: In general, this is the average value of function $f$ over the interval $[a,b]$ : $\dfrac{\int_a^b f(x)\,dx}{b-a}$ In our case, ${f(x)=\sqrt{x}}$, ${a=5}$ and ${b=12}$ : $\begin{aligned} \dfrac{\int_{ a}^{ b} {f(x)}\,dx}{ b- a}&=\dfrac{\int_{{5}}^{ {12}} ({\sqrt{x}})\,dx}{{12}-{5}} \\\\ &=\dfrac{\left[\dfrac{2\cdot x^{^{\frac{3}{2}}}}{3}\right]_{5}^{12}}{7} \\\\ &=\dfrac{\dfrac{2\cdot 12^{^{\frac{3}{2}}}}{3}-\dfrac{2\cdot 5^{^{\frac{3}{2}}}}{3}}{7} \\\\ &=\dfrac{2}{21}\cdot (\sqrt{12^3}-\sqrt{5^3}) \end{aligned}$ In conclusion, the average value of $\sqrt{x}$ on the interval $[5,12]$ is $\dfrac{2}{21}\cdot (\sqrt{12^3}-\sqrt{5^3})$.